The Johnsons and the Wilsons went to the Burger Barn for dinner. The Johnsons bought 2 burger meals and 5 hot dog meals for $46. The Wilsons bought 5 burger meals and 2 hot dog meals for $52. When you solve, you get a solution of (8, 6). (3 points)

Answer 1

To solve for the cost of burger meals and hot dog meals, we can set up a system of equations based on the information provided.

Let:

• \( b \) = cost of a burger meal
• \( h \) = cost of a hot dog meal

From the Johnsons' order, we can create the first equation:

1. \( 2b + 5h = 46 \) (Equation 1)

From the Wilsons' order, we create the second equation: 2. \( 5b + 2h = 52 \) (Equation 2)

Now we can solve this system of equations.

We can use either substitution or elimination; let's use the elimination method.

First, we will multiply Equation 1 by 5 and Equation 2 by 2 to make the coefficients of \( b \) the same:

\[ 5(2b + 5h) = 5(46) \implies 10b + 25h = 230 \quad (Equation 1 \times 5) \]

\[ 2(5b + 2h) = 2(52) \implies 10b + 4h = 104 \quad (Equation 2 \times 2) \]

Now we have:

\[ 10b + 25h = 230 \quad (Equation 3) \] \[ 10b + 4h = 104 \quad (Equation 4) \]

Next, we subtract Equation 4 from Equation 3 to eliminate \( b \):

\[ (10b + 25h) - (10b + 4h) = 230 - 104 \] \[ 21h = 126 \] \[ h = \frac{126}{21} = 6 \]

Now that we have \( h = 6 \), we can substitute \( h \) back into either Equation 1 or Equation 2 to find \( b \). We will use Equation 1:

\[ 2b + 5(6) = 46 \] \[ 2b + 30 = 46 \] \[ 2b = 46 - 30 \] \[ 2b = 16 \] \[ b = \frac{16}{2} = 8 \]

Thus, the solution is: \[ b = 8 \quad \text{(cost of a burger meal)} \] \[ h = 6 \quad \text{(cost of a hot dog meal)} \]

The solution (8, 6) corresponds to \( b = 8 \) and \( h = 6 \).